Question

How do you find the derivative of $f\left( x \right)=\dfrac{1}{{{x}^{2}}}$ using the limit process?

Answer 1

We explain the concept of derivation of a dependent variable with respect to an independent variable. We first find the formula for the derivation for ${{n}^{th}}$ power of a variable x where $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$. We place the value for $n=-2$. We get the solution for the derivative of $f\left( x \right)=\dfrac{1}{{{x}^{2}}}$. We also explain the theorem with the help of the first order derivative.

Complete step by step answer:
Differentiation, the fundamental operations in calculus deals with the rate at which the dependent variable changes with respect to the independent variable. The measurement quantity of its rate of change is known as derivative or differential coefficients. We find the increment of those variables for small changes. We mathematically express it as $\dfrac{dy}{dx}$ where $y=f\left( x \right)$.
The formula of derivation for ${{n}^{th}}$ power of a variable x is
$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$.
The condition being
n\in \mathbb{R} - \left\\{ 0 \right\\}.
For our given function $f\left( x \right)=\dfrac{1}{{{x}^{2}}}$, we can convert it to $f\left( x \right)=\dfrac{1}{{{x}^{2}}}={{x}^{-2}}$.
The value of n is $-2$. We apply the theorem and get
$\dfrac{df}{dx}=\dfrac{d}{dx}\left( {{x}^{-2}} \right)=\left( -2 \right){{x}^{-2-1}}$.
Simplifying the equation, we get
$\dfrac{df}{dx}=\dfrac{d}{dx}\left( {{x}^{-2}} \right)=-2{{x}^{-3}}$. It can also be written as $-2{{x}^{-3}}=\dfrac{-2}{{{x}^{3}}}$.
Therefore, the derivative of the function
$f\left( x \right)=\dfrac{1}{{{x}^{2}}}$ is $\dfrac{-2}{{{x}^{3}}}$.

Note: If the ratio of $\dfrac{\Delta y}{\Delta x}$ tends to a definite finite limit when $\Delta x\to 0$, then the limiting value obtained by this can also be found by first order derivative. We can also apply a first order derivative theorem to get the differentiated value of ${{x}^{\dfrac{3}{4}}}$.
We know that $\dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$. Here $f\left( x \right)={{x}^{n}}$. Also, $f\left( x+h \right)={{\left( x+h \right)}^{n}}$. We assume $x+h=u$ which gives $f\left( u \right)={{\left( u \right)}^{n}}$ and $h=u-x$. As $h\to 0$ we get $u\to x$.